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Elitmus previous papers set - 3

1. There is a equilateral triangle whose area is A. Another triangle is formed by joining the mid points. This process is continued. Find the sum of these areas.
Explanation:

Let ABC is an equilateral triangle with area = A
Remember: The area of the triangle formed when mid points of the triangle is joined is exactly one-fourth of the bigger triangle. So Area of DEF = A/4. Similarly Area of GHI = A/16 and so on ...
So Sum of the areas of all triangles = $A + \dfrac{A}{4} + \dfrac{A}{{16}} + ...$
$\Rightarrow A \left( {1 + \dfrac{1}{4} + \dfrac{1}{{16}} + ...} \right)$
The series in the bracket is G.P. and ${S_\infty } = \dfrac{a}{{1 - r}}$
$ \Rightarrow A\left( \vcenter{\dfrac{1}{{1 - \dfrac{1}{4}}}} \right)$ $ \Rightarrow \dfrac{4}{3}A$

4. 1,2,3,4,5,6,7 are arranged such that sum of two successive numbers is a prime number. For example, 1234765 (i.e. 1+2=3, 2+3=5, 3+4=7....)1. How many such possible combinations occur?2. How many possible combination occurs if first number is 1/7 and last number is 7/1 (i.e 1xxxxx7 or 7xxxxx1)?3. How many numbers will come on 4th position(xxx_xxx)?
Explanation:
We have to follow a systematic approach for this question. We know that if at all two consecutive numbers be prime, they should not be even in the first place. So we arrange even numbers in even places, odd numbers in odd places.
2 should not have 7 as its neighbor
4 should not have 5 as its neighbor
6 should not have 3 as its neighbor
Case 1:
__ 2 ____ 4 ____ 6 __
x7 x7,5 x5,3 x3
So fix, 3 or 1 in between 2 and 4. We get the following options with reverse case also.
5234167, 7614325
5234761, 1674325
1234765, 5674321
3214765, 5674123

5. Find the remainder when 4*4! + 5*5! + 6*6! +...................+ 19*19! is divided by 64.1) 322) 383) 404) Cannot be determined.
Explanation:
8*8! and its subsequent terms are exactly divisible by 64. As highest power of 2 in 64 is 6 and highest power of 2 in 8! is 7 which is greater than 6. So we have to find the remaidner when 4*4! + 5*5! + 6*6! + 7*7! is divided by 64 which is 40.